Difference between revisions of "2019 AIME II Problems/Problem 9"
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− | For <math>n</math> to | + | For <math>n</math> to have exactly <math>20</math> positive divisors, <math>n</math> can only take on certain prime factorization forms: namely, <math>p^{19}, p^9q, p^4q^3, p^4qr</math>. No number that is a multiple of <math>20</math> can be expressed in the first form, and the only integer divisible by <math>20</math> that has the second form is <math>2^{9}5</math>, which is greater than <math>2019</math>. |
− | For the third form, the only numbers | + | For the third form, the only <math>20</math>-pretty numbers are <math>2^45^3=2000</math> and <math>2^35^4=5000</math>, and only <math>2000</math> is small enough. |
− | For the fourth form, <math>2^45r</math> is the | + | For the fourth form, any number of the form <math>2^45r</math> where <math>r</math> is a prime other than <math>2</math> or <math>5</math> will satisfy the <math>20</math>-pretty requirement. Since <math>n=80r<2019</math>, <math>r\le 25</math>. Therefore, <math>r</math> can take on <math>3, 7, 11, 13, 17, 19,</math> or <math>23</math>. |
Thus, <math>\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}</math>. | Thus, <math>\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}</math>. |
Revision as of 17:53, 4 June 2020
Contents
Problem 9
Call a positive integer -pretty if has exactly positive divisors and is divisible by . For example, is -pretty. Let be the sum of positive integers less than that are -pretty. Find .
Solution
Every 20-pretty integer can be written in form , where , , , and , where is the number of divisors of . Thus, we have , using the fact that the divisor function is multiplicative. As must be a divisor of 20, there are not many cases to check.
If , then . But this leads to no solutions, as gives .
If , then or . The first case gives where is a prime other than 2 or 5. Thus we have . The sum of all such is . In the second case and , and there is one solution .
If , then , but this gives . No other values for work.
Then we have .
-scrabbler94
Solution 2
For to have exactly positive divisors, can only take on certain prime factorization forms: namely, . No number that is a multiple of can be expressed in the first form, and the only integer divisible by that has the second form is , which is greater than .
For the third form, the only -pretty numbers are and , and only is small enough.
For the fourth form, any number of the form where is a prime other than or will satisfy the -pretty requirement. Since , . Therefore, can take on or .
Thus, .
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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